 Ballistic orbit insertion with constant gravity

In the flat-surface, constant-gravitational acceleration model, a coasting projectile conserves horizontal velocity. If it arrives at the top of its trajectory with horizontal speed $v_o$, its earlier velocity at angle $\theta$ above the horizontal has vertical component $v_o\tan \theta$. Its remaining height gain is therefore $v_o^2{\tan }^2\theta /(2g)$. The displayed burnout height follows. This is a local projectile motion approximation, not an exact Kepler orbit. Computing $v_o=\sqrt{g(R_E+h_o)}$ assumes constant gravitational magnitude also at the orbit; computing $v_o=\sqrt{g_0{R}_E^2/(R_E+h_o)}$ instead assumes an inverse-square law with surface value $g_0$. Those assumptions must be distinguished.